\newproblem{lay:5_2_9}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.2.9}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Find the characteristic equation and the real eigenvalues of the matrix $A=\begin{pmatrix}4 & 0 & -1 \\ 0 & 4 & -1 \\ 1 & 0 & 2\end{pmatrix}$.
}{
   % Solution
	The characteristic equation is $|A-\lambda I|=0$. In this particular case
	\begin{center}
		$\left|\begin{array}{ccc}4-\lambda & 0 & -1 \\ 0 & 4-\lambda & -1 \\ 1 & 0 & 2-\lambda\end{array}\right|=0$
	\end{center}
	We calculate this determinant by expanding the factors and cofactors of the second column. Disregarding the cofactors multiplied by a zero value, we have
	\begin{center}
		$(4-\lambda)\left|\begin{array}{cc}4-\lambda & -1 \\ 1 & 2-\lambda\end{array}\right|=0$ \\
		$(4-\lambda)((4-\lambda)(2-\lambda)+1)=0$ \\
	\end{center}
	Now, we factorize the term $(4-\lambda)(2-\lambda)+1$
	\begin{center}
		$(4-\lambda)(2-\lambda)+1=(8+\lambda^2-6\lambda)+1=\lambda^2-6\lambda+9=(\lambda-3)^2$ \\
	\end{center}
	So the characteristic equation is
	\begin{center}
		$(4-\lambda)(\lambda-3)^2=0$ \\
	\end{center}
	whose solutions are $\lambda=4$ and $\lambda=3$ (with multiplicity 2).
}
\useproblem{lay:5_2_9}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
